1
Question
A 5.00kg block is placed on top of a 10.0kg block (Fig. above). A horizontal force of 45.0N is applied to the 10kg block, and the 5.00kg block is tied to the wall. The coefficient of kinetic friction between all moving surfaces is 0.200. (a) Draw a free-body diagram for each block and identify the action–reaction forces between the blocks. (b) Determine the tension in the string and the magnitude of the acceleration of the 10.0kg block.
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Solution
(a) The free-body diagrams are shown in the figure below.
f1 and n1 appear in both diagrams as action-reaction pairs.
(b) For the 5.00kg mass, Newton’s second law in the y direction gives:
n1=m1g=(5.00kg)(9.80m/s2)=49.0N
In the x direction,
f1−T=0
T=f1=μmg=0.200(5.00kg)(9.80m/s2)=9.80N
For the 10.0kg mass, Newton’s second law in the x direction gives:
45.0N−f1−f2=(10.0kg)a
In the y direction,
n2−n1−98.0N=0
f2=μn2=μ(n1+98.0N)=0.20(49.0N+98.0N)=29.4N
45.0N−9.80N−29.4N=(10.0kg)a
a=0.580m/s2
f1 and n1 appear in both diagrams as action-reaction pairs.
(b) For the 5.00kg mass, Newton’s second law in the y direction gives:
n1=m1g=(5.00kg)(9.80m/s2)=49.0N
In the x direction,
f1−T=0
T=f1=μmg=0.200(5.00kg)(9.80m/s2)=9.80N
For the 10.0kg mass, Newton’s second law in the x direction gives:
45.0N−f1−f2=(10.0kg)a
In the y direction,
n2−n1−98.0N=0
f2=μn2=μ(n1+98.0N)=0.20(49.0N+98.0N)=29.4N
45.0N−9.80N−29.4N=(10.0kg)a
a=0.580m/s2
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